| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | De Moivre to derive tan/cot identities |
| Difficulty | Challenging +1.3 This is a structured Further Maths question with clear signposting through multiple parts. Part (i) is a standard de Moivre application to derive a triple angle formula. Parts (ii) and (iii) follow guided steps with substitutions provided. While it requires facility with complex numbers, trigonometric identities, and integration techniques, the scaffolding makes it more accessible than questions requiring independent problem-solving or novel insights. |
| Spec | 4.02l Geometrical effects: conjugate, addition, subtraction4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos 3\theta + i \sin 3\theta = c^3 + 3ic^2s - 3cs^2 - is^3\) | M1 | For using de Moivre with \(n = 3\) |
| \(\Rightarrow \cos 3\theta = c^3 - 3cs^2\) and \(\sin 3\theta = 3c^2s - s^3\) | A1 | For both expressions in this form (seen or implied). SR For expressions found without de Moivre M0, A0 |
| \(\Rightarrow \tan 3\theta = \frac{3c^2s - s^3}{c^3 - 3cs^2}\) | M1 | For expressing \(\frac{\sin 3\theta}{\cos 3\theta}\) in terms of \(c\) and \(s\) |
| \(\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \frac{\tan\theta(3 - \tan^2\theta)}{1 - 3\tan^2\theta}\) | A1, 4 | For simplifying to AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta = \frac{\pi}{12} \Rightarrow \tan 3\theta = 1\) | ||
| \(\Rightarrow 1 - 3t^2 = (3 - t^2) \Rightarrow t^3 - 3t^2 - 3t + 1 = 0\) | B1, 1 | For both stages correct AG |
| Answer | Marks | Guidance |
|---|---|---|
| \((t + 1)(t^2 - 4t + 1) = 0\) | M1, A1 | For attempt to factorise cubic. For correct factors |
| \(\Rightarrow (t = -1), t = 2 \pm \sqrt{3}\) | A1 | For correct roots of quadratic |
| \(-\) sign for smaller root \(\Rightarrow \tan\frac{\pi}{12} = 2 - \sqrt{3}\) | A1, 4 | For choice of \(-\) sign and correct root AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(dt = (1 + t^2) d\theta\) | B1 | For differentiation of substitution and use of \(\sec^2\theta = 1 + \tan^2\theta\) |
| \(\Rightarrow \int_0^{\frac{\pi}{12}} \tan 3\theta \, d\theta\) | B1 | For integral with correct \(\theta\) limits seen |
| \(= \left[\frac{1}{3}\ln(\sec 3\theta)\right]_0^{\frac{\pi}{12}} - \frac{1}{3}\ln\left(\sec\frac{\pi}{4}\right)\) | M1 | For integrating to \(k\ln(\sec 3\theta)\) OR \(k\ln(\cos 3\theta)\) |
| \(= \frac{1}{3}\ln\sqrt{2} = \frac{1}{6}\ln 2\) | M1 | For substituting limits and \(\sec\frac{\pi}{4} = \sqrt{2}\) OR \(\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) seen |
| A1, 5 | For correct answer aef |
**Part (i)**
| $\cos 3\theta + i \sin 3\theta = c^3 + 3ic^2s - 3cs^2 - is^3$ | M1 | For using de Moivre with $n = 3$ |
|---|---|---|
| $\Rightarrow \cos 3\theta = c^3 - 3cs^2$ and $\sin 3\theta = 3c^2s - s^3$ | A1 | For both expressions in this form (seen or implied). SR For expressions found without de Moivre M0, A0 |
| $\Rightarrow \tan 3\theta = \frac{3c^2s - s^3}{c^3 - 3cs^2}$ | M1 | For expressing $\frac{\sin 3\theta}{\cos 3\theta}$ in terms of $c$ and $s$ |
| $\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \frac{\tan\theta(3 - \tan^2\theta)}{1 - 3\tan^2\theta}$ | A1, 4 | For simplifying to AG |
**Part (ii)(a)**
| $\theta = \frac{\pi}{12} \Rightarrow \tan 3\theta = 1$ | | |
|---|---|---|
| $\Rightarrow 1 - 3t^2 = (3 - t^2) \Rightarrow t^3 - 3t^2 - 3t + 1 = 0$ | B1, 1 | For both stages correct AG |
**Part (ii)(b)**
| $(t + 1)(t^2 - 4t + 1) = 0$ | M1, A1 | For attempt to factorise cubic. For correct factors |
|---|---|---|
| $\Rightarrow (t = -1), t = 2 \pm \sqrt{3}$ | A1 | For correct roots of quadratic |
| $-$ sign for smaller root $\Rightarrow \tan\frac{\pi}{12} = 2 - \sqrt{3}$ | A1, 4 | For choice of $-$ sign and correct root AG |
**Part (iii)**
| $dt = (1 + t^2) d\theta$ | B1 | For differentiation of substitution and use of $\sec^2\theta = 1 + \tan^2\theta$ |
|---|---|---|
| $\Rightarrow \int_0^{\frac{\pi}{12}} \tan 3\theta \, d\theta$ | B1 | For integral with correct $\theta$ limits seen |
|---|---|---|
| $= \left[\frac{1}{3}\ln(\sec 3\theta)\right]_0^{\frac{\pi}{12}} - \frac{1}{3}\ln\left(\sec\frac{\pi}{4}\right)$ | M1 | For integrating to $k\ln(\sec 3\theta)$ OR $k\ln(\cos 3\theta)$ |
| $= \frac{1}{3}\ln\sqrt{2} = \frac{1}{6}\ln 2$ | M1 | For substituting limits and $\sec\frac{\pi}{4} = \sqrt{2}$ OR $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ seen |
| | A1, 5 | For correct answer aef |
---7
\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to prove that
$$\tan 3 \theta \equiv \frac { \tan \theta \left( 3 - \tan ^ { 2 } \theta \right) } { 1 - 3 \tan ^ { 2 } \theta } .$$
\item (a) By putting $\theta = \frac { 1 } { 12 } \pi$ in the identity in part (i), show that $\tan \frac { 1 } { 12 } \pi$ is a solution of the equation
$$t ^ { 3 } - 3 t ^ { 2 } - 3 t + 1 = 0 .$$
(b) Hence show that $\tan \frac { 1 } { 12 } \pi = 2 - \sqrt { 3 }$.
\item Use the substitution $t = \tan \theta$ to show that
$$\int _ { 0 } ^ { 2 - \sqrt { 3 } } \frac { t \left( 3 - t ^ { 2 } \right) } { \left( 1 - 3 t ^ { 2 } \right) \left( 1 + t ^ { 2 } \right) } \mathrm { d } t = a \ln b$$
where $a$ and $b$ are positive constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2009 Q7 [14]}}